c++ - C++中的值比较结果

Question

在下面的代码中，输出为1，但我不确定为什么。我是否已隐式更改为 bool(boolean) 值？如何将整数值设置为这样的表达式？

#include <iostream>
using namespace std;

int main (int argc, const char *argv[])
{
  int a = 1, b = 1, c = 1, i = 1;
  i = b < a < c;
  cout << i;
  return 0;
}

Answer 1

Has i been implicitly changed to a boolean?

不会。相反， bool(boolean) 表达式已隐式转换为 int，然后存储在 i中，如@WhozCraig所评论。

How can an integer value be set to an expression like this?

因为它遵循C++标准。

当您这样做时:

i = b < a < c;

由于第一个 <和第二个 <具有相同的运算符优先级(因为它们是相同的运算符)，因此:

Operators that have the same precedence are bound to their argumentsin the direction of their associativity. For example, the expression a= b = c is parsed as a = (b = c), and not as (a = b) = c because of right-to-left associativity of assignment, but a + b - c is parsed (a > + b) - c and not a + (b - c) because of left-to-right associativity of addition and subtraction.

如 C++ Operator Precedence中所述，这意味着将解析您的表达式，就像它是这样写的:

i = (b < a) < c;

这将评估括号内的条件，即False。
现在 bool到 int的转换是隐式的:
C++标准的第4.7 / 4节说(整体转换)

If the source type is bool, the value false is converted to zero and the value true is converted to one.

这意味着将false转换为0，然后我们基本上完成了:

i = 0 < c;

由于 c等于1，其结果为true。现在 i的类型为 int，这意味着发生了另一次第二次从 bool(boolean) 值到整数的隐式转换，最终将1分配给 i。